A “descending number” is a three-digit number, such that the units digit is less than the tens digits and the tens digit is less than the hundreds digit. What is the probability that a three-digit number chosen at random is a “descending number”?

(A) 3/25

(B) 2/9

(C) 2/15

(D) 1/9

(E) 1/10


Kudos for correct solution.

The total 3 digit numbers can be formed by FCP i.e = 9*10*10 (we have digit from 0 -9 and there are no restriction then the hundred digit can be from 1-9,as it cannot start from 0 , tens digit can be 0-9 and units can also be 0-9).

Now we have to choose the three digit numbers in descending order -

We can leave out 100 - 200 = as they donot form in descending order

Next from 200 to 200 = we have only one number =210

Next from 300 to 400 = we have 3 nos = 310,320,321

Next from 400 to 500 = we have 6 nos = 410,420,421,430,431,432

Now we find that it is following a pattern and i.e



200 = 1

300 = 3 (2+1 from previous possible ways , 2 is the hundred unit )

400 = 6 (3+3)

500 = 10 (4+6,similarly 4 is the hundred digit and 6 in the value from the previous)

600 = 15(5+10)

700 = 21 (6+15)

800 = 28 (7+21)

900 = 36(8+28)

Therefore no. of possible ways = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 =120

Therefore \(probability = \frac{120}{900} = \frac{2}{15}\)
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